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Chapter 4: Q27E (page 913)

Crossed E and B Fields. A particle with initial velocityv鈫�=(5.85脳103m/s)j^enters a region of uniform electric and magnetic fields. The magnetic field in the region isB鈫�=-(1.35T)k. Calculate the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge (a) +0.640 nC and (b) -0.320 nC. You can ignore the weight of the particle.

Short Answer

Expert verified

a) The magnitude and direction of the charged particle of charge +0.640 nC is (7898N/C)i^

b) The magnitude and direction of the charged particle of charge -0.320 nC is (-7898N/C)i^

Step by step solution

01

Magnetic force and electric force

The magnetic force on a particle is given by

FB鈫�=q(v鈫�脳B鈫�)

Here q is the charge of the particle and v is the velocity of the particle

And the electric force is

FE鈫�=qE鈫�

02

Determine the magnitude and direction of the first particle

(a)

The magnetic force is

FB鈫�=q(v鈫�脳B鈫�)

And the electric force on the particle is

FE鈫�=qE鈫�

Both forces are equal

Therefore,

FE鈫�=FB鈫�qvB=qEE=vB

Put the values in the equation

E=(5.85脳103m/s)(1.35T)=7898N/C

And the direction is j^脳(-k)=-i^

Therefore, the magnitude and direction of the charged particle of charge +0.640 nC is(7898N/C)i^

03

Determine the direction and magnitude of the second particle

(b)

Same as the previous one,

The electric field of the second particle is

E鈫�=vB(鈭�i^)E=(5.85脳103m/s)(1.35T)(鈭�i^)=鈭�7898N/C

Therefore, the magnitude and direction of the charged particle of charge +0.640 nC is(鈭�7898N/C)i^

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